Question

solve for the variable in each figure.

1. ( y = )

( x = )

1. ( k = )
2. ( sq = x ); ( st = 22 );

( sp = 12 ); ( pr = 4x + 8 )
( x = )

1. ( y = )

( x = )

Explanation:

Step1: Use the basic proportionality theorem (Thales' theorem) for the first figure

If a line is parallel to one side of a triangle and intersects the other two sides, then it divides those sides proportionally.
For the sides with \(y\): \(\frac{5}{5 + 2}=\frac{5y-2}{5y - 2+3x + 2}\), but also for the parallel lines cutting the sides: \(\frac{5}{2}=\frac{5y-2}{10}\) (cross - multiply)
\(5\times10=2\times(5y - 2)\)
\(50 = 10y-4\)
\(10y=54\)
\(y=\frac{54}{10}=5.4\)
For \(x\): \(\frac{5}{2}=\frac{10}{x}\) (cross - multiply)
\(5x=20\)
\(x = 4\)

Step2: Use the basic proportionality theorem for the second figure

\(\frac{k}{4}=\frac{9}{12}\) (cross - multiply)
\(12k=9\times4\)
\(12k = 36\)
\(k = 3\)

Step3: Use the basic proportionality theorem for the third figure (assuming the line is parallel)

\(\frac{SQ}{ST}=\frac{SP}{SP + PR}\), given \(SQ=x\), \(ST = 22\), \(SP = 12\), \(PR=4x + 8\)
\(\frac{x}{22}=\frac{12}{12+(4x + 8)}\)
\(\frac{x}{22}=\frac{12}{4x+20}\)
Cross - multiply: \(x(4x + 20)=12\times22\)
\(4x^{2}+20x=264\)
\(x^{2}+5x - 66=0\)
Factor: \((x + 11)(x - 6)=0\)
\(x=- 11\) (rejected as length \(\gt0\)) or \(x = 6\)

Step4: Use the basic proportionality theorem for the fourth figure

For \(y\): \(\frac{8}{y}=\frac{16}{20}\) (cross - multiply)
\(16y=8\times20\)
\(16y = 160\)
\(y = 10\)
For \(x\): Let the sides of the triangle be \(a\) and \(b\). Using the ratio \(\frac{8}{8 + y}=\frac{x}{x + 5}\), since \(y = 10\)
\(\frac{8}{18}=\frac{x}{x + 5}\)
Cross - multiply: \(8(x + 5)=18x\)
\(8x+40=18x\)
\(10x = 40\)
\(x = 4\)

Answer:

1. \(y = 5.4\), \(x = 4\)
2. \(k = 3\)
3. \(x = 6\)
4. \(y = 10\), \(x = 4\)